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3 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

1 answer:

4 0

Every science experiment should follow the basic principles of proper investigation so that the results presented at the end are seen as credible.

Observation and Hypothesis. ...

Prediction and Modeling. ...

Testing and Error Estimation. ...

Result Gathering and Presentation. ...

Conclusions. ...

Law Formation.

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Exercise 3: Alkynes A. For the following alkyne structure, state the hybridization of each carbon in the molecule and predict th

Lerok [7]

For the following alkyne structure, the hybridization of each carbon in the molecule is C1 = Sp3, C2=Sp3, C3=Sp, C4=Sp, C5= Sp3

Bond angle=

1)C1-C2-C3= 109 degree 28

2)C2-C3-C4=180 degree

3)C3-C4-C5= 180 degrees

Effective molecular charge order= S> Si> AI> Mg

What is hybridization?

In chemistry, orbital hybridization (or hybridization) is the concept of mixing atomic orbitals to form new hybrid orbitals (having different energies, shapes, etc. than the component atomic orbitals) suitable for electron pairing. Yes, they form chemical bonds through valence bond formation.

Hybridization is defined as the concept of mixing two atomic orbitals to create a new type of hybrid orbital. This mixing usually results in the formation of hybrid orbitals with very different energies, shapes, etc.

To know more about hybridization, click here- brainly.com/question/22765530

#SPJ4

5 0

1 year ago

What is the molarity of chloride ion in a solution made by dissolving 1.300g of aluminumchloride in a total volume of 500.0mL?

navik [9.2K]

Answer:

0.0582 M

Explanation:

The following data were obtained from the question:

Mass of AlCl3 = 1.3 g

Volume of water = 500 mL

Molarity of chloride ion (Cl¯) =?

Next, we shall determine the number of mole in 1.3 g of AlCl3. This can be obtained as follow:

Mass of AlCl3 = 1.3 g

Molar mass of AlCl3 = 27 + (35.5×3)

= 27 + 106.5

= 133.5 g/mol

Mole of AlCl3 =?

Mole = mass /molar mass

Mole of AlCl3 = 1.3/133.5

Mole of AlCl3 = 0.0097 mole

Next, we shall convert 500 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Thus, 500 mL is equivalent to 0.5 L.

Next, we shall determine the molarity of the AlCl3 solution. This can be obtained as follow:

Mole of AlCl3 = 0.0097 mole

Volume of water = 0.5 L

Molarity of AlCl3 =?

Molarity = mole /Volume

Molarity of AlCl3 = 0.0097 / 0.5

Molarity of AlCl3 = 0.0194 M

Next, we shall write the dissociation equation of AlCl3 in solution. This is illustrated below:

AlCl3 (aq) —> Al³⁺ (aq) + 3Cl¯ (aq)

From the balanced equation above,

1 mole of AlCl3 produced 3 moles of Cl¯.

Finally, we shall determine the molarity of the chloride ion Cl¯ in the solution as follow:

From the balanced equation above,

1 mole of AlCl3 produced 3 moles of Cl¯.

Therefore, 0.0194 M AlCl3 will produce = 0.0194 × 3 = 0.0582 M Cl¯.

Thus, the molarity of the chloride ion, Cl¯ in the solution is 0.0582 M.

8 0

3 years ago

The mixture you separated was a mixture of iron filings, sand, and salt. Based on your understanding of matter, is this mixture

Anarel [89]

Its a heterogeneous mixture

The parts are not uniformly mixed

5 0

3 years ago

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t

Tamiku [17]

Answer:

The explosive decomposition of 98.4 kg of ammonium nitrate produces 58498.8 L of nitrogen, 29249.4 L of oxygen and 116997.6 L of water vapor.

Explanation:

To find how many liters of gas are formed from the explosive decomposition of ammonium nitrate it is necessary to follow these steps (check the attachment for better understanding):

1st) Balance the equation:

Write the decomposition equation and then find the correct coefficients to make sure that it goes according to the "Law of conservation of mass" (the mass of the reactants side must be equal to the mass of the products side). So, 2 moles of ammonium nitrate produces 2 moles of nitrogen, 1 mole of oxygen and 4 moles of water vapor.

2nd) Find the Ammonium nitrate molar mass:

The ammonium nitrate mass it is calculated by adding de molar mass of each atom that forms the ammonium nitrate molecule. You can find the elements molar mass in the Periodic Table.

In this example I decided to round the number to simplify tha calculus, for example: the oxygen molar mass in the periodic table is 15.9994 but I use 16. You can use the complete number if you want.

By doing this, the ammonium nitrate molar mass is 80 g/mol.

The statement says that there is 98.4 kg of ammonium nitrate. In ordder to use the same units in all the calculus sometimes it is usefull to convert the kg to g, so it is the same as 98400g. You can do it the other way around if you prefer (g to kg).

3rd) Find the number of moles of each gases and aqua vapor formed:

It is important to know the amount of each compound formed by the decomposition reaction, that's why we need to pay attention to the coefficients of the balanced reaction.

The amount of each compound is easily found by using the "rule of three".

To use the rule of three we need to think using the balanced reaction so:

If 160g (2 moles) of ammonium nitrate produces 2 moles of nitrogen gas, the 98400g that we have of ammonium nitrate will produce an X amount of nitrogen gas. With this information we multiply 98400g by 2 moles and then we divide the result by 160g. The final result it is 1230 moles of nitrogen.

In the same way we use the rule of three to calculate the number of moles of oxygen and water.

4th) Find the liters (volume) of each gas and aqua vapor formed:

Finally, to find the liters from the number of moles, it is necessary to apply the "Ideal gases law", that relates the pressure (atm), volume (L), moles number and temperature (Kelvin) with the R gas constant in the formula:

PxV = nxRxT

It is important to use the correct units because the R gas constant is equal to 0.082 atm.L/mol.K.

As we need to calculate the liters (volume) we pass the pressure dividing to the other side and then we just have to replace the information:

V = (nxRxT)/P

As you can see in the attachment, doing this last step for each compound, we can find the liters produced of them.

8 0

3 years ago

In the following reaction, how many liters of oxygen produce 560 liters of

saw5 [17]

The balanced chemical reaction will be:

CH4 + 2O2 → CO2 + 2H2O

We are given the amount of carbon dioxide to produce from the reaction. This will be our starting point.

560 L CH4 ( 1 mol CH4/ 22.4 L CH4 ) (2 mol O2/ 1 mol CH4 ) ( 22.4 L O2 / 1 mol <span>O2</span><span>) = 1120 L O2</span>

7 0

3 years ago

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